$\begin{array}{1 1}1196 \\ 1210 \\ 1312 \\ 1218 \end{array} $

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- $n^{th}$ term of an A.P.=$a+(n-1)d$
- Sum of $n$ terms of an A.P=$\large\frac{n}{2}$$(t_n+a)$

Two digit numbers that yields remainder $1$ when divided by $4$ forms sequence

$13,17,21,........97$

This sequence is an A.P with first term $a=13$ common difference $d=4$ and

$t_n=97$

$\Rightarrow\:a+(n-1)d=97$

$\Rightarrow\:13+(n-1)4=97$

$\Rightarrow\:n-1=\large\frac{84}{4}$$=21$

or $n=22$

$\therefore$ Sum of the required numbers =$13+17+21+.......97$

$=\large\frac{n}{2}$$(t_n+a)=11(97+13)=11\times 110=1210$

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